# Showing a Hölder continuous function acted on by a singular integral operator is Hölder continuous

Consider the following function defined by a singular integral

$$F(x)= \lim_{\epsilon \rightarrow 0} \int_{|x-y| \geq \epsilon} \partial_k \partial_j k_i(x-y) \left(Y_k(x)- Y_k(y) \right)f(y)dy$$

where $f \in C_c^{\gamma}(\mathbb{R}^n)$ (ie a compactly supported $\gamma$-Hölder continuous function) and $k$ is defined by $$k_i(x) =\frac{x_i}{|x|^n}$$ We also have $Y:\mathbb{R}^n \rightarrow \mathbb{R}^n$ such that \begin{equation*} \| Y\|_{1,\gamma} \equiv |Y(0)| +\| \nabla Y\|_{sup} +|\nabla Y|_{\gamma} < \infty \end{equation*} where $$|W|_{\gamma} = \sup_{x \neq y} \frac{|W(x)-W(y)|}{|x-y|^{\gamma}}$$ (ie $Y$ is a $C^1(\mathbb{R}^n,\mathbb{R}^n)$ function with bounded and $\gamma$-Hölder continuous derivatives).

I was wondering how to prove $F$ is $\gamma$-Hölder continuous. I have tried writing out $Y$ as the sum of a first order Taylor polynomial and its corresponding remainder term. I did this to make use of the following property of the kernel $k$:

$$\int_{|x|=1}x^{\beta}D^{\alpha}k_i(x)dS=0 \quad \Rightarrow \int_{r \leq |x|\leq R} x^{\beta}D^{\alpha}k_i(x) dx = 0$$ for any multi-indices $\alpha$ and $\beta$ such that $|\alpha|=|\beta|+1$. In particular, terms of the form $$C\int_{r \leq |x|\leq R} x_l \partial_k\partial_jk_i(x) dx = 0$$ can be added freely to help with the inequality for the first order $Y$ term.

I'd also be interested in knowing whether this kind of thing is treated in a textbook somewhere.

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